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THEOREM XII.

If in any triangle ABC the side AB be greater than the side AC, then shall the angle C be greater than the angle B.

Let AD be the perpendicular. Then since AB is greater than AC, it is more remote from the perpendicular.

A

B C'

Hence if ADC be turned about AD, C will fall between D and B, as at C'.

Then the exterior angle AC'D is greater than ABD; that is, C> B.

Conversely, if the angle C be greater than the angle B, then shall AB be greater than AC.

COR. If the angle C be equal to the angle B, then shall AB be equal to AC.

Congruent Triangles.

Every triangle has six parts, viz. the three sides and the three angles. When the six parts of one triangle are equal to the six parts of another, each to each, the triangles are said to be congruent.

We shall show that two triangles must be congruent if in the one the following parts are equal to the corresponding parts of the other.

I. The three sides.

II. Two sides and the included angle.

III. Two angles and a corresponding side.

We shall also investigate the case of two triangles, which have two sides of the one, and an angle not the included angle, equal to the corresponding parts of the other.

THEOREM XIII.

If in the triangles ABC, A'B'C',

AB=A'B', BC= B'C', CA = C'A',

then shall CC', A= A', and B = B'.

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Apply the triangle ABC to the triangle A'B'C' so that the point B may be on B', and BC on B'C'; C will coincide with C' since BC= B'C'. Join AA'.

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Therefore whole (fig. 1) or rem. (fig. 2) B'AC' = whole or

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AB= A'B', BC= B'C', and B=B',

then shall CA = C'A'; C = C', and A = A'.

Apply the triangle ABC to the triangle A'B'C', so that A may be on A and AB fall on A'B'.

Then because AB = A'B',

B will coincide with B'.

And, because the angle B = the angle B',
BC will take the direction of B'C'.

And, because BC= B'C',

the point C will coincide with the point C'.

Therefore CA must coincide with C'A', and be equal to it.

And the angle C must coincide with the angle C', and be equal to it.

And the angle A must coincide with the angle A', and be equal to it.

THEOREM XV.

If in the triangles ABC, A'B'C',

B=B', C=C', and BC= B'C',

then shall A = A', CA = C'A', and AB=A'B'.

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To prove the rest of the proposition, apply the triangle ABC to the triangle A'B'C', so that B may be on B', and BC have the direction of B'C'.

Then, since BC= B'C',

C must coincide with C'.

Now, since BB',

BA must take the direction of B'A',

and the point A must lie somewhere on B'A'.

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CA must take the direction of C'A',

and the point A must lie somewhere on C'A'.

Therefore A must coincide with A'.

Therefore CA must coincide with C'A', and be equal

to it.

And AB must coincide with A'B', and be equal to it.

Ambiguous Case.

In the triangles ABC, A'B'C', let

CA = C'A', AB=A'B', B=B'.

A A A

B

C B'

Apply the triangle ABC to the triangle A'B'C', so that A may be on A', and AB have the direction of A'B'.

Then, since AB=A'B',

the point B must coincide with B'.

And since BB',

BC must take the direction of B'C'.

From A draw A'D perpendicular to B'C' produced, if necessary, and take DE=DC'.

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these lines must be equally remote from the perpendicular A'D.

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